A circular disc of radius b has a hole of radius a at its centre(see figure). If the mass per unit area of…

A circular disc of radius $b$ has a hole of radius $\mathrm{a}$ at its centre(see figure). If the mass per unit area of the disc varies as $\left(\frac{{\sigma }_0}{r}\right)$ then, the radius of gyration of the disc about its axis passing through the center is 

1. $\frac{a+b}{3}$
2. $\sqrt{\frac{a^2+b^2+ab}{3}}$
3. $\frac{a+b}{2}$
4. $\sqrt{\frac{a^2+b^2+ab}{2}}$

Solution

Solution: (B)
$dm=\sigma 2\pi rdr$
$=\frac{{\sigma }_0}{r}2\pi rdr$
$m=2\pi {\sigma }_0{\int }_a^{b}dr=2\pi \hspace{0.17em}{\sigma }_0(b-a)$

$I=2\pi {\sigma }_0{\int }_a^b{r}^{2}dr=\frac{2}{3}\pi {\sigma }_0\left(b^3-a^3\right)$
Radius of gyration $=\sqrt{\frac{I}{m} }=\sqrt{\frac{\frac{2}{3}\pi {\sigma }_0\left(b^3-a^3\right)}{2\pi {\sigma }_0\left(b-a\right)}}$
$=\sqrt{\frac{b^3-a^3}{3(b-a)}}=\sqrt{\frac{(b-a)(b^2+a^2+ab)}{3(b-a)}}$
$=\sqrt{\frac{b^2+a^2+ab}{3}}$. 

Asked in: JEE Main 2019 (12 Apr Shift 1)

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